Question

What is the lattice energy of calcium chloride?

Answer 1

Lattice energy

1. Lattice energy is defined as the energy released during the conversion of gaseous ions into their corresponding solid ions.
2. Calcium chloride has cation ${\left(\mathrm{Ca}\right)}^{2+}$and two anions of Chlorine${\left(\mathrm{Cl}\right)}^{-}$.
3. The lattice energy of Calcium chloride $\left({\mathrm{CaCl}}_2\right)$is$-2195.2kJmo{l}^{-}$.

Born-Haber Cycle

1. Born-Haber cycle is the cycle in which a change in enthalpy leads to the formation of Ionic compounds from the elemental atoms and the enthalpy of formation of solid compounds so that the net enthalpy becomes zero.
2. Empirically we cannot calculate the lattice energy but it can be measured using electrostatics or the Born Haber cycle.
3. Born Haber cycle is an approach to analyzing reaction energies.
4. Born Haber cycle for the formation of${\mathrm{CaCl}}_2$.
   $\mathrm{Sublimation}:\mathrm{Ca}\left(\mathrm{s}\right)\to \mathrm{Ca}\left(\mathrm{g}\right)=∆{\mathrm{H}}_{1°}$$\mathrm{Ionization}:\mathrm{Ca}\left(\mathrm{g}\right)\to {\mathrm{Ca}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}=∆{\mathrm{H}}_{2°}$$\mathrm{Vaporisation}:{\mathrm{Cl}}_2\left(\mathrm{l}\right)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)=∆{\mathrm{H}}_{3°}$$\mathrm{Dissociation}:{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to 2\mathrm{Cl}\left(\mathrm{g}\right)=∆{\mathrm{H}}_{4°}$$\mathrm{Electron}\mathrm{affinity}:2\mathrm{Cl}\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)=∆{\mathrm{H}}_{5°}$$\mathrm{Lattice}\mathrm{enthalpy}:{\mathrm{Ca}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)\to {\mathrm{CaCl}}_2\left(\mathrm{s}\right)=∆{\mathrm{H}}_{6°}$$∆{\mathrm{H}}_{{\mathrm{f}}^{\circ }}=∆{\mathrm{H}}_{1°}+∆{\mathrm{H}}_{2°}+∆{\mathrm{H}}_{3°}+∆{\mathrm{H}}_{5°}+∆{\mathrm{H}}_{6°}$
   By knowing these values we can calculate the lattice energy of Calcium chloride.

Calculation of $\left({\mathrm{CaCl}}_2\right)$Lattice energy

Formula used- $$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{f}}^{°}=\mathrm{Lattice}\mathrm{energy}+\left(2\times \mathrm{electron}\mathrm{affinity}\mathrm{of}\mathrm{Chlorine}\right)+\mathrm{Bond}\mathrm{energy}\mathrm{of}\mathrm{Chlorine}\mathrm{gas} \\ +\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{ionization}\mathrm{energy}\mathrm{of}\mathrm{Ca}+\mathrm{Sublimation}\mathrm{energy}\mathrm{of}\mathrm{Ca}\left(\mathrm{s}\right) \end{gathered}$$

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{sublimation}}=121\mathrm{KJ} \\ 1\mathrm{st}\mathrm{ionization}\mathrm{energy}\mathrm{of}\mathrm{Ca}=589.5\mathrm{KJ} \\ 2\mathrm{nd}\mathrm{ionization}\mathrm{energy}\mathrm{of}\mathrm{Ca}=1145\mathrm{KJ} \\ \mathrm{Bond}\mathrm{energy}\mathrm{of}{\mathrm{Cl}}_2=242.7\mathrm{KJ} \\ \mathrm{Electron}\mathrm{affinity}\mathrm{of}\mathrm{Cl}=2\times \left(-349\mathrm{KJ}\right) \\ \mathrm{Standard}\mathrm{enthalpy}\mathrm{of}\mathrm{formation}\mathrm{of}\mathrm{CaCl}=-795\mathrm{KJ} \end{gathered}$$

By substituting the values in the formula we get

$$\begin{gathered} -795\mathrm{KJ}=\mathrm{Lattice}\mathrm{energy}+2\times -349\mathrm{KJ}+242.4\mathrm{KJ}+1145\mathrm{KJ}+589.5\mathrm{KJ} \\ \mathrm{Lattice}\mathrm{energy}=-795\mathrm{KJ}-2\times \left(-349\mathrm{KJ}\right)-242.4\mathrm{KJ}-1145\mathrm{KJ}-589.5\mathrm{KJ}-121\mathrm{KJ} \\ \mathrm{Lattice}\mathrm{energy}=-2195.5{\mathrm{KJmol}}^{-} \end{gathered}$$